Question

If $A$ is the orthocentre of the triangle formed by $2 x^{2}-y^{2}=0, x+y-1=0$ and $B$ is the centroid of the triangle formed by $2 x^{2}-5 x y+2 y^{2}=0,7 x-2 y-12=0$, then the distance between $A$ and $B$ is

• A. $\sqrt{5}$
• B. 1
• C. 5
• D. $\sqrt{2}$

Answer 1

Answer: (A)

Given,
$2 x^{2}-y^{2}=0 $
$\Rightarrow y^{2}=2 x^{2}$
$\Rightarrow y=\pm \sqrt{2} x$ and $x+y=1$

$\because$ Point of intersection of perpendicular is orthocentre.
The two perpendiculars are $y=x$ and $x-\sqrt{2} y+\sqrt{2}-1=0$
$\therefore $ Point of intersection of $y=x$ and $x-\sqrt{2} y+\sqrt{2}-1=0$
$\therefore A=(1,1)$
Now, $2 x^{2}-5 x y+2 y^{2}=0$
$\Rightarrow (2 x-y)(y-2 x)=0$
$\Rightarrow y=2 x$ or $x=2 y$ and $7 x-2 y-12=0$

$\therefore $ Centroid $=\left(\frac{0+4+2}{3}, \frac{0+8+1}{3}\right)\left(\frac{6}{3}, \frac{9}{3}\right)=(2,3)=B$
Now, required distance
$=\sqrt{1+4}=\sqrt{5}$